What is the integral from 0 to 4 of (3t)/((t-5)^2) ?

The integral from 0 to 4 of (3t)/((t-5)^2) is 3(-ln(5)+4)

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integral from 0 to 4 of (3t)/((t-5)^2)

\int_{0}^{4} \frac{3 t}{( t - 5 ) ^{2}} d t

3 (- ln (5) + 4)

Decimal Notation

7.17168 \dots

Solution steps

\int_{0}^{4} \frac{3 t}{( t - 5 ) ^{2}} d t

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 3 \cdot \int_{0}^{4} \frac{t}{( t - 5 ) ^{2}} d t

Apply u-substitution

: \int_{- 5}^{- 1} \frac{u + 5}{u ^{2}} d u

= 3 \cdot \int_{- 5}^{- 1} \frac{u + 5}{u ^{2}} d u

Expand

\frac{u + 5}{u ^{2}} : \frac{1}{u} + \frac{5}{u ^{2}}

= 3 \cdot \int_{- 5}^{- 1} \frac{1}{u} + \frac{5}{u ^{2}} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 3 (\int_{- 5}^{- 1} \frac{1}{u} d u + \int_{- 5}^{- 1} \frac{5}{u ^{2}} d u)

\int_{- 5}^{- 1} \frac{1}{u} d u = - ln (5)

\int_{- 5}^{- 1} \frac{5}{u ^{2}} d u = 4

= 3 (- ln (5) + 4)

f (x) = \frac{x}{x ^{2} + 6}

x \to 2 lim (5 x^{3} - 9 x^{2} + 1)

x \to \infty lim (4^{x} x)

\frac{3 s ^{2}}{( s ^{2} + 1 ) ^{2}}

\int y \cdot e^{x \cdot y} d x

What is the integral from 0 to 4 of (3t)/((t-5)^2) ?
The integral from 0 to 4 of (3t)/((t-5)^2) is 3(-ln(5)+4)